Existence results for impulsive nonlinear fractional differential equation with mixed boundary conditions

Abstract

In this paper, the existence and uniqueness of solutions for an impulsive mixed boundary value problem of nonlinear differential equations of fractional order are obtained. Our results are based on some fixed point theorems. Some examples are also presented to illustrate the main results.

Introduction

Recently, boundary value problems of nonlinear fractional differential equations have been addressed by several researchers. Fractional differential equations arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in the fields of physics, chemistry, control theory, biology, economics, blood flow phenomena, signal and image processing, biophysics, aerodynamics, fitting of experimental data, etc. For details, see  and the references therein.

Impulsive differential equations, which provide a natural description of observed evolution processes, are regarded as important mathematical tools for a better understanding of several real world problems in the applied sciences. Recently, the boundary value problems of impulsive differential equations of integer order have been studied extensively in the literature (see [1, 37, 913]). In [4, 13], Wang et al. gave a new concept of some impulsive differential equations with fractional derivative, which is a correction of that of piecewise continuous solutions used in [3, 7, 1012].

This paper is strongly motivated by the above research papers. We investigate the existence and uniqueness of solutions for a mixed boundary value problem of nonlinear impulsive differential equations of fractional order given by

$$\left \{ \textstyle\begin{array}{@{}l} ^{C}D_{0^{+}}^{q} u(t)=f(t,u(t)), \quad t \in J', \\ \Delta u(t_{k})=I_{k}(u(t_{k})),\qquad \Delta u'(t_{k})=J_{k}(u(t_{k})),\quad k=1, 2, \ldots, p, \\ u(0) + u'(0)=0, \qquad u(1) + u'(1)=0, \end{array}\displaystyle \right .$$
(1.1)

where $$^{C}D_{0^{+} }^{q}$$ is the Caputo fractional derivative of order $$q \in(1, 2)$$, $$f\in C(J\times R, R)$$. $$I_{k}, J_{k}\in C(R, R)$$, $$J=[0, 1]$$, $$J'=J\setminus\{t_{1}, t_{2}, \ldots, t_{p}\}$$, the $$\{t_{k}\}$$ satisfy $$0=t_{0}< t_{1}< t_{2}<\cdots<t_{p}<t_{p + 1}=1$$, $$p\in N$$, $$\Delta u(t_{k}) = u(t_{k}^{+} ) - u(t_{k}^{-} )$$, $$\Delta u'(t_{k})=u'(t_{k}^{+} ) - u'(t_{k}^{-} )$$, where $$u(t_{k}^{+} )$$ and $$u(t_{k}^{-} )$$ represent the right and left limits of $$u(t)$$ at $$t=t_{k}$$.

A function $$u \in PC(J, R)$$ is said to be a solution of problem (1.1) if $$u(t) = u_{k}(t)$$ for $$t \in(t_{k}, t_{k+1})$$ and $$u_{k} \in C([0, t_{k+1}], R)$$ satisfies $$^{C}D_{0^{+}}^{q} u(t)=f(t,u(t))$$ a.e. on $$(0, t_{k+1})$$ with the restriction that $$u_{k}(t)$$ on $$[0, t_{k})$$ is just $$u_{k-1}(t)$$ and the conditions $$\Delta u(t_{k})=I_{k}(u(t_{k}))$$, $$\Delta u'(t_{k})=J_{k}(u(t_{k}))$$, $$k=1, 2, \ldots, p$$ with $$u(0) + u'(0)=0$$, $$u(1) + u'(1)=0$$.

The rest of this paper is organized as follows. In Section 2, we give some notations, recall some concepts and preparation results. In Section 3, we give the main results, the first result based on Banach contraction principle, the second result based on Krasnoselskii’s fixed point theorem. Two examples are given in Section 4 to demonstrate the application of our main results.

Preliminaries

In this section, we introduce preliminary facts which are used throughout this paper.

Let $$J_{0}=[0, t_{1}], J_{1}=(t_{1}, t_{2}], \ldots, J_{p - 1}=(t_{p - 1}, t_{p}], J_{p}=(t_{p}, 1]$$. We have

\begin{aligned} PC(J) =&\bigl\{ u: [0,1] \rightarrow R\mid u \in C\bigl(J' \bigr), \mbox{ and } u\bigl(t_{k}^{+} \bigr), u\bigl(t_{k}^{-} \bigr) \mbox{ exist, and}\\ &{}u\bigl(t_{k}^{-} \bigr)=u(t_{k}), 1\leq k\leq p\bigr\} . \end{aligned}

Obviously, $$PC(J)$$ is a Banach space with the norm

$$\|u\|_{PC}=\sup_{0\le t\le1}\bigl|u(t)\bigr|.$$

Definition 2.1

The fractional integral of order q of a function $$f: [0, \infty) \to R$$ is defined as

$$I_{0 + }^{q} f(t)=\frac{1}{\Gamma(q)} \int_{0}^{t} \frac{f(s)}{(t - s)^{1 - q}}\,ds,\quad t>0, q>0,$$
(2.1)

provided the right side is point-wise defined on $$(0, \infty)$$, where $$\Gamma(\cdot)$$ is the gamma function.

Definition 2.2

The Caputo derivative of fractional order q for a function $$f: [0, \infty) \rightarrow R$$ is defined as

\begin{aligned}& {}^{C}D_{0^{+}}^{q} f(t)=\frac{1}{\Gamma(n - q)} \frac{d^{n}}{dt^{n}} \int_{0}^{t} \frac {f(s)-\sum_{k=0}^{n-1}\frac{s^{k}}{k!}f^{(k)}(0)}{(t - s)^{q - n + 1}}\,ds, \quad t>0, n=-[-q], \end{aligned}
(2.2)

where $$[q]$$ denotes the integer part of the real number q.

Remark 2.1

In the case $$f(t) \in C^{n}[0, +\infty)$$, there is $$^{C}D_{0^{+}}^{q} f(t) = I_{0+}^{n-q}f^{(n)}(t)$$. That is to say that Definition 2.2 is just the usual Caputo’s fractional derivative. In this paper, we consider an impulsive problem, so Definition 2.2 is appropriate.

Lemma 2.1

()

Let M be a closed, convex, and nonempty subset of a Banach space X, and A, B the operators such that

1. (1)

$$Ax + By\in M$$ whenever $$x, y\in M$$;

2. (2)

A is compact and continuous;

3. (3)

B is a contraction mapping.

Then there exists $$z\in M$$ such that $$z=Az + Bz$$.

Lemma 2.2

()

The set $$F \subset PC([0,T], R^{n})$$ is relatively compact if and only if:

1. (i)

F is bounded, that is, $$\|x\|\leq C$$ for each $$x\in F$$ and some $$C>0$$;

2. (ii)

F is quasi-equicontinuous in $$[0,T]$$. That is to say that for any $$\epsilon>0$$ there exists $$\delta>0$$ such that if $$x \in F$$; $$k \in N$$; $$\tau_{1}, \tau_{2} \in(t_{k - 1}, t_{k}]$$, and $$|\tau_{1} - \tau_{2}|<\delta$$, we have $$|x(\tau_{1}) - x(\tau_{2})|<\epsilon$$.

Lemma 2.3

()

For $$q>0$$, the general solution of the fractional differential equation $$^{C}D_{0^{+}}^{q} u(t)=0$$ is given by

$$u(t)=c_{0} + c_{1} t + c_{2}t^{2} + \cdots + c_{n - 1}t^{n - 1},$$

where $$c_{i} \in R$$, $$i=0, 1, 2, \ldots, n - 1$$, $$n=-[-q]$$.

In view of Lemma 2.3, it follows that

$$I_{0 + }^{q} \bigl({}^{C}D_{0^{+}}^{q} u\bigr) (t) = u(t) + c_{0} + c_{1}t + c_{2}t^{2} + \cdots + c_{n - 1}t^{n - 1},$$

where $$c_{i} \in R$$, $$i=0, 1, 2, \ldots, n - 1$$, $$n=-[-q]$$.

Lemma 2.4

Let $$q\in(1, 2)$$ and $$h: J\rightarrow R$$ be continuous. A function u given by

$$u(t)= \textstyle\begin{cases} \frac{1}{\Gamma(q)}\int_{0}^{t} (t - s)^{q - 1}h(s)\,ds + \frac{1 - t}{\Gamma (q)}\int_{0}^{1}(1 - s)^{q - 1}h(s)\,ds\\ \quad{}+ \frac{1-t}{\Gamma(q - 1)}\int_{0}^{1}(1 - s)^{q - 2}h(s)\,ds, \quad t\in [0,t_{1}]; \\ \frac{1}{\Gamma(q)}\int_{0}^{t} (t - s)^{q - 1}h(s)\,ds + \frac{1-t}{\Gamma (q)}\int_{0}^{1}(1 - s)^{q - 1}h(s)\,ds\\ \quad{}+ \frac{1-t}{\Gamma(q - 1)}\int_{0}^{1}(1 - s)^{q - 2}h(s)\,ds + (2 - t) \sum_{j=1}^{p} J_{j} (u(t_{j} ))(1 - t_{j} )\\ \quad{}+ (2 - t)\sum_{j=1}^{p} I_{j} (u(t_{j} )) - (t - t_{j} )\sum_{j=k + 1}^{p} J_{j} (u(t_{j} )) - \sum_{j=k + 1}^{p} I_{j} (u(t_{j} )),\\ \quad t\in(t_{k}, t_{k + 1}], k=1, 2, \ldots, p - 1;\\ \frac{1}{\Gamma(q)}\int_{0}^{t} (t - s)^{q - 1}h(s)\,ds + \frac{1-t}{\Gamma (q)}\int_{0}^{1}(1 - s)^{q - 1}h(s)\,ds\\ \quad{}+ \frac{1-t}{\Gamma(q - 1)}\int_{0}^{1}(1 - s)^{q - 2}h(s)\,ds + (2 - t) \sum_{j=1}^{p} J_{j} (u(t_{j} ))(1 - t_{j} )\\ \quad{}+ (2 - t)\sum_{j=1}^{p} I_{j} (u(t_{j} )), \quad t\in(t_{p}, t_{p + 1}], \end{cases}$$
(2.3)

is a unique solution of the following impulsive problem:

$$\textstyle\begin{cases} {}^{C}D_{0^{+}}^{q} u(t)=h(t), \quad t \in J',\\ \Delta u(t_{k})=I_{k}(u(t_{k})), \qquad \Delta u'(t_{k})=J_{k}(u(t_{k})),\quad k=1, 2, \ldots, p, \\ u(0) + u'(0)=0, \qquad u(1) + u'(1)=0. \end{cases}$$
(2.4)

Proof

With Lemma 2.3, a general solution u of the equation $$^{C}D_{0^{+}}^{q} u(t)=h(t)$$ on each interval $$(t_{k}, t_{k + 1}]$$ ($$k=0, 1, 2, \ldots, p$$) is given by

$$u(t)=\frac{1}{\Gamma(q)} \int_{0}^{t} (t - s)^{q - 1}h(s)\,ds + a_{k} + b_{k} t, \quad\mbox{for } t\in(t_{k}, t_{k + 1}],$$
(2.5)

where $$t_{0}=0$$ and $$t_{p + 1}=1$$. Then we have

$$u'(t)=\frac{1}{\Gamma(q - 1)} \int_{0}^{t} (t - s)^{q - 2}h(s)\,ds + b_{k}, \quad\mbox{for } t\in(t_{k}, t_{k + 1}].$$
(2.6)

We have

\begin{aligned}& u(0)=a_{0}, \qquad u'(0)=b_{0},\\& u(1)=\frac{1}{\Gamma(q)} \int_{0}^{1} (1 - s)^{q - 1}h(s)\,ds + a_{p} + b_{p},\\& u'(1)=\frac{1}{\Gamma(q - 1)} \int_{0}^{1} (1 - s)^{q - 2}h(s)\,ds + b_{p}. \end{aligned}

So applying the boundary conditions (2.4), we have

\begin{aligned}& a_{0} + b_{0}=0, \end{aligned}
(2.7)
\begin{aligned}& \frac{1}{\Gamma(q)} \int_{0}^{1} (1 - s)^{q - 1}h(s)\,ds + \frac{1}{\Gamma(q - 1)} \int_{0}^{1} (1 - s)^{q - 2}h(s)\,ds + a_{p} + 2b_{p}=0. \end{aligned}
(2.8)

Furthermore, using the impulsive condition $$\Delta u'(t_{k})=u'(t_{k}^{+} ) - u'(t_{k}^{-} )=J_{k}(u(t_{k}))$$, we derive

\begin{aligned}& b_{k}=b_{k - 1} + J_{k}\bigl(u(t_{k}) \bigr), \end{aligned}
(2.9)
\begin{aligned}& b_{k} = b_{p} - \sum_{j=k + 1}^{p} J_{j}\bigl(u(t_{j})\bigr) \quad(k=1, 2, \ldots, p - 1). \end{aligned}
(2.10)

In the same way, using the impulsive condition $$\Delta u(t_{k})=u(t_{k}^{+} ) - u(t_{k}^{-} )=I_{k}(u(t_{k}))$$, we derive

$$a_{k} + b_{k} t_{k} = a_{k - 1} + b_{k - 1}t_{k} + I_{k} \bigl(u(t_{k})\bigr),$$
(2.11)

which by (2.9) implies that

$$a_{k} = a_{k - 1} - J_{k}\bigl(u(t_{k}) \bigr)t_{k} + I_{k}\bigl(u(t_{k})\bigr).$$
(2.12)

Thus

$$a_{k} = a_{p} + \sum_{j=k + 1}^{p} J_{j}\bigl(u(t_{j})\bigr)t_{j} - \sum _{j=k + 1}^{p} I_{j}\bigl(u(t_{j}) \bigr)\quad (k=0, 1, 2, \ldots, p - 1).$$
(2.13)

Combining (2.7), (2.8), (2.10) with (2.13) yields

\begin{aligned}& \begin{aligned}[b] a_{p}={}&\frac{1}{\Gamma(q)} \int_{0}^{1}(1 - s)^{q - 1}h(s)\,ds + \frac{1}{\Gamma (q - 1)} \int_{0}^{1} (1 - s)^{q - 2}h(s)\,ds\\ &{} - 2\sum_{j=1}^{p} J_{j} \bigl(u(t_{j})\bigr) (t_{j} - 1) + 2\sum _{j=1}^{p} I_{j}\bigl(u(t_{j}) \bigr), \end{aligned} \end{aligned}
(2.14)
\begin{aligned}& \begin{aligned}[b] b_{p}= {}& {-} \frac{1}{\Gamma(q)} \int_{0}^{1}(1 - s)^{q - 1}h(s)\,ds - \frac {1}{\Gamma(q - 1)} \int_{0}^{1}(1 - s)^{q - 2}h(s)\,ds\\ &{} + \sum_{j=1}^{p} J_{j} \bigl(u(t_{j} )\bigr) (t_{j} - 1) - \sum _{j=1}^{p} I_{j} \bigl(u(t_{j} ) \bigr). \end{aligned} \end{aligned}
(2.15)

Furthermore, by (2.10), (2.13), (2.14), (2.15) we have

\begin{aligned}& \begin{aligned}[b] a_{k} ={}&\frac{1}{\Gamma(q)} \int_{0}^{1}(1 - s)^{q - 1}h(s)\,ds + \frac{1}{\Gamma (q - 1)} \int_{0}^{1}(1 - s)^{q - 2}h(s)\,ds\\ &{}- 2\sum_{j=1}^{p} J_{j} \bigl(u(t_{j} )\bigr) (t_{j} - 1) + 2\sum _{j=1}^{p} I_{j} \bigl(u(t_{j} ) \bigr)\\ &{} + \sum_{j=k + 1}^{p} J_{j} \bigl(u(t_{j} )\bigr)t_{j} - \sum _{j=k + 1}^{p} I_{j} \bigl(u(t_{j} ) \bigr) \quad(k=0, 1, 2, \ldots, p-1), \end{aligned} \end{aligned}
(2.16)
\begin{aligned}& \begin{aligned}[b] b_{k} ={}& {-} \frac{1}{\Gamma(q)} \int_{0}^{1}(1 - s)^{q - 1}h(s)\,ds - \frac {1}{\Gamma(q - 1)} \int_{0}^{1}(1 - s)^{q - 2}h(s)\,ds\\ &{}+ \sum_{j=1}^{p} J_{j} \bigl(u(t_{j} )\bigr) (t_{j} - 1) - \sum _{j=1}^{p} I_{j} \bigl(u(t_{j} ) \bigr)\\ &{}- \sum_{j=k + 1}^{p} J_{j} \bigl(u(t_{j} )\bigr) \quad(k=0, 1, 2, \ldots, p-1). \end{aligned} \end{aligned}
(2.17)

Hence for $$k=0, 1, 2, \ldots, p-1$$, (2.16) and (2.17) imply

\begin{aligned} a_{k} + b_{k} t={}&\frac{1-t}{\Gamma(q)} \int_{0}^{1}(1 - s)^{q - 1}h(s)\,ds + \frac{1-t}{\Gamma(q - 1)} \int_{0}^{1}(1 - s)^{q - 2}h(s)\,ds \\ &{} + (2 - t)\sum_{j=1}^{p} J_{j} \bigl(u(t_{j} )\bigr) (1 - t_{j} ) + (2 - t)\sum _{j=1}^{p} I_{j} \bigl(u(t_{j} ) \bigr) \\ &{}- (t - t_{j} )\sum_{j=k + 1}^{p} J_{j} \bigl(u(t_{j} )\bigr) - \sum _{j=k + 1}^{p} I_{j} \bigl(u(t_{j} ) \bigr). \end{aligned}
(2.18)

For $$k=p$$, (2.14) and (2.15) imply

\begin{aligned} a_{k} + b_{k} t={}&\frac{1-t}{\Gamma(q)} \int_{0}^{1}(1 - s)^{q - 1}h(s)\,ds + \frac{1-t}{\Gamma(q - 1)} \int_{0}^{1}(1 - s)^{q - 2}h(s)\,ds \\ &{} + (2 - t)\sum_{j=1}^{p} J_{j} \bigl(u(t_{j} )\bigr) (1 - t_{j} ) + (2 - t)\sum _{j=1}^{p} I_{j} \bigl(u(t_{j} ) \bigr). \end{aligned}
(2.19)

Now it is clear that (2.5), (2.18), (2.19) imply that (2.3) holds.

Conversely, assume that u satisfies (2.3). By a direct computation, it follows that the solution given by (2.3) satisfies (2.4). □

Main results

This section deals with the existence and uniqueness of solutions to problem (1.1).

Theorem 3.1

Let $$f: J\times R\rightarrow R$$ be a continuous function. Suppose there exist positive constants $$L_{1}$$, $$L_{2}$$, $$L_{3}$$, $$M_{2}$$, $$M_{3}$$ such that

1. (A1)

$$|f(t,x) - f(t,y)|\leq L_{1} |x - y|$$, for all $$t\in J$$, $$x, y\in R$$;

2. (A2)

$$|I_{k} (x) - I_{k} (y)|\leq L_{2} |x - y|$$, $$|J_{k} (x) - J_{k} (y)|\leq L_{3}|x - y|$$, $$|I_{k} (x)|\leq M_{2}$$, $$|J_{k}(x)|\leq M_{3}$$, $$x, y\in R$$, $$k=1, 2, \ldots, p$$,

with

$$L_{1} \leq\frac{\Gamma(q + 1)}{2(2 + q)},\qquad L_{1} \biggl[ \frac{2}{\Gamma(q + 1)} + \frac{1}{\Gamma(q)} \biggr] + 3p(L_{2} + L_{3})< 1.$$

Then problem (1.1) has a unique solution on J.

Proof

Define an operator $$T: PC(J)\rightarrow PC(J)$$

\begin{aligned} (Tu) (t) :={}&\frac{1}{\Gamma(q)} \int_{0}^{t} (t - s)^{q - 1}f\bigl(s,u(s) \bigr)\,ds + \frac{1-t}{\Gamma(q)} \int_{0}^{1}(1 - s)^{q - 1}f\bigl(s,u(s) \bigr)\,ds \\ &{} + \frac{1-t}{\Gamma(q - 1)} \int_{0}^{1}(1 - s)^{q - 2}f\bigl(s,u(s) \bigr)\,ds + (2 - t)\sum_{j=1}^{p} J_{j} \bigl(u(t_{j} )\bigr) (1 - t_{j} ) \\ &{} + (2 - t)\sum_{j=1}^{p} I_{j} \bigl(u(t_{j} )\bigr) - (t - t_{j} )\sum _{j=k + 1}^{p} J_{j} \bigl(u(t_{j} ) \bigr) - \sum_{j=k + 1}^{p} I_{j} \bigl(u(t_{j} )\bigr), \\ &{} t\in(t_{k} ,t_{k + 1}], k=0, 1, 2, \ldots, p. \end{aligned}

Let $$\sup_{t\in J}|f(t,0)|=M$$, and $$B_{r} =\{u\in PC(J,R)\mid\|u\| _{PC}\leq r\}$$, where

$$r\geq2 \biggl[\frac{2 + q}{\Gamma(q + 1)}M + 3p(M_{2} + M_{3}) \biggr].$$

Step 1. We show that $$TB_{r} \subset B_{r}$$.

For $$u\in B_{r}$$, $$t\in J$$, we have

\begin{aligned} &\bigl|(Tu) (t)\bigr|\\ &\quad \leq\frac{1}{\Gamma(q)} \int_{0}^{t} (t - s)^{q - 1}\bigl|f\bigl(s,u(s) \bigr)\bigr|\,ds + \frac {1}{\Gamma(q)} \int_{0}^{1}(1 - s)^{q - 1}\bigl|f\bigl(s,u(s) \bigr)\bigr|\,ds\\ &\qquad{} + \frac{1}{\Gamma(q - 1)} \int_{0}^{1}(1 - s)^{q - 2}\bigl|f\bigl(s,u(s) \bigr)\bigr|\,ds + 2\sum_{j=1}^{p} \bigl|J_{j} \bigl(u(t_{j} )\bigr)\bigr|\\ &\qquad{} + 2\sum_{j=1}^{p} \bigl|I_{j} \bigl(u(t_{j} )\bigr)\bigr| + \sum_{j=k + 1}^{p} \bigl|J_{j} \bigl(u(t_{j} )\bigr)\bigr| + \sum _{j=k + 1}^{p} \bigl|I_{j} \bigl(u(t_{j} )\bigr)\bigr|\\ &\quad\leq\frac{1}{\Gamma(q)} \biggl[ \int_{0}^{t} (t - s)^{q - 1}\bigl|f\bigl(s,u(s) \bigr) - f(s,0)\bigr|\,ds + \int_{0}^{t} (t - s)^{q - 1}\bigl|f(s,0)\bigr|\,ds \biggr]\\ &\qquad{} + \frac{1}{\Gamma(q)} \biggl[ \int_{0}^{1}(1 - s)^{q - 1}\bigl|f\bigl(s,u(s) \bigr) - f(s,0)\bigr|\,ds + \int_{0}^{1}(1 - s)^{q - 1}\bigl|f(s,0)\bigr|\,ds \biggr]\\ &\qquad{} + \frac{1}{\Gamma(q - 1)} \int_{0}^{1}(1 - s)^{q - 2}\bigl|f\bigl(s,u(s) \bigr) - f(s,0)\bigr|\,ds\\ &\qquad{} +\frac{1}{\Gamma(q - 1)} \int_{0}^{1}(1 - s)^{q - 2}\bigl|f(s,0)\bigr|\,ds + 2\sum _{j=1}^{p} \bigl|J_{j} \bigl(u(t_{j} )\bigr)\bigr|\\ &\qquad{} + 2\sum_{j=1}^{p} \bigl|I_{j} \bigl(u(t_{j} )\bigr)\bigr| + \sum_{j=k + 1}^{p} \bigl|J_{j} \bigl(u(t_{j} )\bigr)\bigr| + \sum _{j=k + 1}^{p} \bigl|I_{j} \bigl(u(t_{j} )\bigr)\bigr|\\ &\quad\leq\frac{L_{1} r}{\Gamma(q + 1)} + \frac{M}{\Gamma(q + 1)} + \frac{L_{1} r}{\Gamma(q + 1)} + \frac{M}{\Gamma(q + 1)} + \frac{L_{1} r}{\Gamma(q)} + \frac{M}{\Gamma(q)} \\ &\qquad{} + 2pM_{3} + 2pM_{2} + pM_{3} + pM_{2} \\ &\quad=L_{1} \frac{2 + p}{\Gamma(q + 1)}r + \frac{2 + p}{\Gamma(q + 1)}M + 3p(M_{2} + M_{3}). \end{aligned}

Since

$$L_{1} \leq\frac{\Gamma(q + 1)}{2(2 + q)},\qquad r\geq2 \biggl[\frac{2 + q}{\Gamma(q + 1)}M + 3p(M_{2} + M_{3}) \biggr],$$

we have

$$\bigl|(Tu) (t)\bigr|\leq r,\quad TB_{r} \subset B_{r} .$$

Step 2. T is a contraction mapping.

For $$x, y\in B_{r}$$ and $$t\in J$$, we have

\begin{aligned} &\bigl|(Tx) (t) - (Ty) (t)\bigr|\\ &\quad=\Biggl\vert \frac{1}{\Gamma(q)} \int_{0}^{t} (t - s)^{q - 1}f\bigl(s,x(s) \bigr)\,ds + \frac{1-t}{\Gamma(q)} \int_{0}^{1}(1 - s)^{q - 1}f\bigl(s,x(s) \bigr)\,ds \\ &\qquad{} + \frac{1-t}{\Gamma(q - 1)} \int_{0}^{1}(1 - s)^{q - 2}f\bigl(s,x(s) \bigr)\,ds + (2 - t)\sum_{j=1}^{p} J_{j} \bigl(x(t_{j} )\bigr) (t_{j} - 1)\\ &\qquad{} + (2 - t)\sum_{j=1}^{p} I_{j} \bigl(x(t_{j} )\bigr) - (t - t_{j} )\sum _{j=k + 1}^{p} J_{j} \bigl(x(t_{j} ) \bigr) - \sum_{j=k + 1}^{p} I_{j} \bigl(x(t_{j} )\bigr)\\ &\qquad{} - \Biggl[\frac{1}{\Gamma(q)} \int_{0}^{t} (t - s)^{q - 1}f\bigl(s,y(s) \bigr)\,ds + \frac{1-t}{\Gamma(q)} \int_{0}^{1}(1 - s)^{q - 1}f\bigl(s,y(s) \bigr)\,ds \\ &\qquad{} +\frac{1-t}{\Gamma(q - 1)} \int_{0}^{1}(1 - s)^{q - 2}f\bigl(s,y(s) \bigr)\,ds + (2 - t)\sum_{j=1}^{p} J_{j} \bigl(y(t_{j} )\bigr) (t_{j} - 1)\\ &\qquad{} + (2 - t)\sum_{j=1}^{p} I_{j} \bigl(y(t_{j} )\bigr) - (t - t_{j} )\sum _{j=k + 1}^{p} J_{j} \bigl(y(t_{j} )\bigr) - \sum_{j=k + 1}^{p} I_{j} \bigl(y(t_{j} )\bigr) \Biggr]\Biggr\vert \\ &\quad\leq\frac{1}{\Gamma(q)} \int_{0}^{t} (t - s)^{q - 1}\bigl|f\bigl(s,x(s) \bigr) - f\bigl(s,y(s)\bigr)\bigr|\,ds\\ &\qquad{} + \frac{1}{\Gamma(q)} \int_{0}^{1}(1 - s)^{q - 1}\bigl|f\bigl(s,x(s) \bigr) - f\bigl(s,y(s)\bigr)\bigr|\,ds\\ &\qquad{} + \frac{1}{\Gamma(q - 1)} \int_{0}^{1}(1 - s)^{q - 2}\bigl|f\bigl(s,x(s) \bigr) - f\bigl(s,y(s)\bigr)\bigr|\,ds\\ &\qquad{} + 2\sum_{j=1}^{p} \bigl|J_{j} \bigl(x(t_{j} )\bigr) - J_{j} \bigl(y(t_{j} ) \bigr)\bigr| + 2\sum_{j=1}^{p} \bigl|I_{j} \bigl(x(t_{j} )\bigr) - I_{j} \bigl(y(t_{j} ) \bigr)\bigr|\\ &\qquad{} + \sum_{j=k + 1}^{p} \bigl|J_{j} \bigl(x(t_{j} )\bigr) - J_{j} \bigl(y(t_{j} ) \bigr)\bigr| + \sum_{j=k + 1}^{p} \bigl|I_{j} \bigl(x(t_{j} )\bigr) - I_{j} \bigl(y(t_{j} ) \bigr)\bigr|\\ &\quad\leq\frac{L_{1} }{\Gamma(q + 1)}\|x - y\|_{PC} + \frac{L_{1} }{\Gamma(q + 1)}\|x - y \|_{PC} + \frac{L_{1} }{\Gamma(q)}\|x - y\|_{PC}\\ &\qquad{} +2\sum_{j=1}^{p} L_{3}\|x - y \|_{PC}+ 2\sum_{j=1}^{p} L_{2} \|x - y\|_{PC}\\ &\qquad{} + \sum_{j=k + 1}^{p} L_{3}\|x - y \|_{PC} + \sum_{k=j + 1}^{p} L_{2} \|x - y\|_{PC}\\ &\quad\leq\frac{2L_{1} }{\Gamma(q + 1)}\|x - y\|_{PC} + \frac{L_{1} }{\Gamma (q)}\|x - y \|_{PC} + 2pL_{3}\|x - y\|_{PC}\\ &\qquad{} + 2pL_{2} \|x - y\|_{PC} + pL_{3}\|x - y \|_{PC} + pL_{2} \|x - y\| _{PC}\\ &\quad= \biggl[L_{1} \biggl(\frac{2}{\Gamma(q + 1)} + \frac{1}{\Gamma(q)}\biggr) + 3p(L_{2} + L_{3}) \biggr]\|x - y\|_{PC}. \end{aligned}

Since

$$L_{1} \biggl(\frac{2}{\Gamma(q + 1)} + \frac{1}{\Gamma(q)} \biggr) + 3p(L_{2} + L_{3})< 1,$$

T is a contraction mapping. Thus, the conclusion follows by the contraction mapping principle. □

Theorem 3.2

Assume that $$|f(t,u)|\leq\mu(t)$$ for all $$(t,u)\in J\times R$$ where $$\mu\in L^{1/\sigma}(J,R)$$ and $$\sigma\in(0, q - 1)$$, furthermore, there exist positive constants $$L_{2}$$, $$L_{3}$$, $$M_{2}$$, $$M_{3}$$ such that $$|I_{k} (x) - I_{k} (y)|\leq L_{2} |x - y|$$, $$|J_{k} (x) - J_{k} (y)|\leq L_{3}|x - y|$$, $$|I_{k} (x)|\leq M_{2}$$, $$|J_{k}(x)|\leq M_{3}$$, $$x, y\in R$$, $$k=1, 2, \ldots, p$$, with $$3p(L_{2} + L_{3})<1$$. Then problem (1.1) has at least one solution on J.

Proof

Choose

$$r\geq\|\mu\|_{L^{\frac{1}{\sigma}}(J)} \biggl[ \frac{2}{\Gamma(q)(\frac {q - \sigma}{1 - \sigma})^{1 - \sigma}} + \frac{1}{\Gamma(q - 1)(\frac{q - \sigma- 1}{1 - \sigma})^{1 - \sigma }} \biggr] + p(M_{2} + 3M_{3})$$

and denote

$$B_{r} =\bigl\{ u\in PC(J,R) \mid \|u\|_{PC}\leq r\bigr\} .$$

Define the operators P and Q on $$B_{r}$$ as

\begin{aligned}& \begin{aligned}[b] (Pu) (t)={}&\frac{1}{\Gamma(q)} \int_{0}^{t} (t - s)^{q - 1}f\bigl(s,u(s) \bigr)\,ds + \frac{1-t}{\Gamma(q)} \int_{0}^{1}(1 - s)^{q - 1}f\bigl(s,u(s) \bigr)\,ds \\ &{} +\frac{1-t}{\Gamma(q - 1)} \int_{0}^{1}(1 - s)^{q - 2}f\bigl(s,u(s) \bigr)\,ds, \end{aligned}\\& \begin{aligned}[b] (Qu) (t)={}&(2 - t)\sum_{k=1}^{p} J_{k} \bigl(u(t_{k} )\bigr) (1 - t_{k} ) + (2 - t) \sum_{k=1}^{p} I_{k} \bigl(u(t_{k} )\bigr)\\ &{} - (t - t_{k} )\sum_{k=j + 1}^{p} J_{k} \bigl(u(t_{k} )\bigr) - \sum _{k=j + 1}^{p} I_{k} \bigl(u(t_{k} ) \bigr). \end{aligned} \end{aligned}

For any $$u,v\in B_{r}$$ and $$t\in J$$, using the condition that $$|f(t,u)|\leq\mu(t)$$ and the Hölder inequality,

\begin{aligned}& \int_{0}^{t} \bigl|(t - s)^{q - 1}f\bigl(s,u(s) \bigr)\bigr|\,ds\\& \quad \leq \biggl( \int_{0}^{t} (t - s)^{\frac{q - 1}{1 - \sigma}}\,ds \biggr)^{1 - \sigma} \biggl( \int_{0}^{t} \bigl(\mu(s)\bigr)^{\frac{1}{\sigma}}\,ds \biggr)^{\sigma} \leq\frac{\|\mu\|_{L^{\frac{1}{\sigma}}(J)}}{(\frac{q - \sigma}{1 - \sigma})^{1 - \sigma}},\\& \int_{0}^{t} \bigl|(1 - s)^{q - 1}f\bigl(s,u(s) \bigr)\bigr|\,ds\\& \quad \leq \biggl( \int_{0}^{t} (1 - s)^{\frac{q - 1}{1 - \sigma}}\,ds \biggr)^{1 - \sigma} \biggl( \int_{0}^{t} \bigl(\mu(s)\bigr)^{\frac{1}{\sigma}}\,ds \biggr)^{\sigma} \leq\frac{\|\mu\|_{L^{\frac{1}{\sigma}}(J)}}{(\frac{q - \sigma}{1 - \sigma})^{1 - \sigma}},\\& \int_{0}^{t} \bigl|(1 - s)^{q - 2}f\bigl(s,u(s) \bigr)\bigr|\,ds \\& \quad \leq \biggl( \int_{0}^{t} (1 - s)^{\frac{q - 2}{1 - \sigma}}\,ds \biggr)^{1 - \sigma} \biggl( \int_{0}^{t} \bigl(\mu(s)\bigr)^{\frac{1}{\sigma}}\,ds \biggr)^{\sigma} \leq\frac{\|\mu\|_{L^{\frac{1}{\sigma}}(J)}}{(\frac{q - \sigma- 1}{1 - \sigma})^{1 - \sigma}}. \end{aligned}

Therefore,

\begin{aligned}& \|Pu + Qv\|_{PC}\\& \quad\leq\frac{2\|\mu\|_{L^{\frac{1}{\sigma}}(J)}}{\Gamma(q)(\frac{q - \sigma}{1 - \sigma})^{1 - \sigma}} + \frac{\|\mu\|_{L^{\frac{1}{\sigma }}(J)}}{\Gamma(q - 1)(\frac{q - \sigma- 1}{1 - \sigma})^{1 - \sigma}}\\& \qquad{}+ 2pM_{3} + 2pM_{2} + pM_{3} + pM_{2}\\& \quad=\|\mu\|_{L^{\frac{1}{\sigma}}(J)} \biggl( \frac{2}{\Gamma(q)(\frac{q - \sigma}{1 - \sigma})^{1 - \sigma}} + \frac{1}{\Gamma(q - 1) (\frac{q - \sigma- 1}{1 - \sigma})^{1 - \sigma }} \biggr) + 3p(M_{2} + M_{3}). \end{aligned}

Thus $$Pu + Qv\in B_{r}$$. It is obvious that Q is a contraction mapping (the proof is just similar to Theorem 3.1). On the other hand, the continuity of f implies that the operator P is continuous. Also, P is uniformly bounded on $$B_{r}$$ since

$$\|Pu\|_{PC}\leq\frac{2\|\mu\|_{L^{\frac{1}{\sigma}}(J)}}{\Gamma (q)(\frac{q - \sigma}{1 - \sigma})^{1 - \sigma}} + \frac{\|\mu\|_{L^{\frac{1}{\sigma}}(J)}}{\Gamma(q - 1)(\frac{q - \sigma- 1}{1 - \sigma})^{1 - \sigma}}\leq r.$$

Now we prove the quasi-equicontinuity of the operator P.

Let $$\Omega=J\times B_{r}$$, $$f_{\max}=\sup_{(t,u)\in\Omega}|f(t,u)|$$. For any $$t_{k} <\tau_{2}<\tau_{1}\leq t_{k + 1}$$, we have

\begin{aligned}& \bigl|(Pu) (\tau_{2}) - (Pu) (\tau_{1})\bigr|\\& \quad= \biggl\vert \frac{1}{\Gamma(q)} \int_{0}^{\tau_{2}}(\tau_{2} - s)^{q - 1}f\bigl(s,u(s)\bigr)\,ds + \frac{1 - \tau_{2}}{\Gamma(q)} \int_{0}^{1}(1 - s)^{q - 1}f\bigl(s,u(s) \bigr)\,ds \\& \qquad{} + \frac{1 - \tau_{2}}{\Gamma(q - 1)} \int_{0}^{1}(1 - s)^{q - 2}f\bigl(s,u(s) \bigr)\,ds - \frac{1}{\Gamma(q)} \int_{0}^{\tau_{1}}(\tau_{1} - s)^{q - 1}f\bigl(s,u(s)\bigr)\,ds\\& \qquad{} - \frac{1 - \tau_{1}}{\Gamma(q)} \int_{0}^{1}(1 - s)^{q - 1}f\bigl(s,u(s) \bigr)\,ds - \frac{1 - \tau_{1}}{\Gamma(q - 1)} \int_{0}^{1}(1 - s)^{q - 2}f\bigl(s,u(s) \bigr)\,ds\biggr\vert \\& \quad\leq\frac{f_{\max}}{\Gamma(q)}\biggl\vert \int_{0}^{\tau_{2}}\bigl[(\tau_{2} - s)^{q - 1} - (\tau_{1} - s)^{q - 1}\bigr]\,ds + \int_{\tau_{2}}^{\tau_{1}}(\tau_{1} - s)^{q - 1}\,ds \biggr\vert \\& \qquad{} + \biggl\vert \frac{(\tau_{1} - \tau_{2})f_{\max}}{\Gamma(q)} \int_{0}^{1}(1 - s)^{q - 1}\,ds\biggr\vert + \biggl\vert \frac{(\tau_{1} - \tau_{2})f_{\max}}{\Gamma(q - 1)} \int_{0}^{1}(1 - s)^{q - 2}\,ds\biggr\vert \\& \quad\leq f_{\max} \biggl[\frac{2(\tau_{1} - \tau_{2})^{q} + \tau_{1}^{q} - \tau_{2}^{q} + \tau_{1} - \tau_{2}}{\Gamma(q + 1)} + \frac{\tau_{1} - \tau_{2}}{\Gamma(q)} \biggr], \end{aligned}

which tends to zero as $$\tau_{2}\rightarrow\tau_{1}$$. This shows that P is quasi-equicontinuous on the interval $$(t_{k} ,t_{k + 1}]$$. It is obvious that P is compact by Lemma 2.2, so P is relatively compact on $$B_{r}$$.

Thus all the assumptions of Lemma 2.1 are satisfied and problem (1.1) has at least one solution on J. □

Example

Example 4.1

Consider the following impulsive fractional boundary value problem:

$$\textstyle\begin{cases} ^{C}D_{0+}^{\frac{3}{2}}u(t)=\frac{1}{(t + 3)^{2}}\frac{\sin^{5} u(t)}{1 + u^{4}(t)}, \quad t \in[0,1], t \neq\frac{1}{4}, \\ \Delta u(\frac{1}{4})=\frac{|u(\frac{1}{4})|}{15 + |u(\frac{1}{4})|}, \qquad \Delta u'(\frac{1}{4})=\frac{|u(\frac{1}{4})|}{17 + |u(\frac {1}{4})|}\\ u(0) + u'(0)=0, \qquad u(1) + u'(1)=0. \end{cases}$$
(4.1)

Obviously, $$L_{1} =1/9$$, $$L_{2} =1/15$$, $$L_{3}=1/17$$, $$M_{2}=1/15$$, $$M_{3}=1/17$$, $$p=1$$,

\begin{aligned}& \frac{\Gamma(q + 1)}{2(2 + q)}=\frac{3\sqrt{\pi}}{28},\qquad L_{1} < \frac{\Gamma (q + 1)}{2(2 + q)}, \\& L_{1} \biggl(\frac{2}{\Gamma(q + 1)} + \frac{1}{\Gamma(q)} \biggr) + 3p(L_{2} + L_{3})=\frac{14}{27\sqrt{\pi}} + \frac{96}{255}< 1. \end{aligned}

Thus, all the assumptions in Theorem 3.1 are satisfied. Hence, the impulsive fractional boundary value problem (4.1) has a unique solution on $$[0,1]$$.

Example 4.2

Consider the following impulsive fractional boundary value problem:

$$\textstyle\begin{cases} {}^{C}D_{0+}^{\frac{3}{2}}u(t)=\frac{e^{t}}{(t + 1)^{2}}\frac{|u(t)|}{1 + |u(t)|}, \quad t \in[0, 1], t \neq\frac{1}{4}, \\ \Delta u(\frac{1}{4})=\frac{3 + 2|u(\frac{1}{4})|}{1 + |u(\frac{1}{4})|}, \qquad\Delta u'(\frac{1}{4})=\frac{5|u(\frac{1}{4})|}{1 + |u(\frac {1}{4})|}\\ u(0) + u'(0)=0, \qquad u(1) + u'(1)=0. \end{cases}$$
(4.2)

Set

$$f(t,u)=\frac{e^{t}}{(t + 1)^{2}}\frac{|u|}{1 + |u|},\quad (t, u)\in[0,1]\times [0,\infty).$$

Obviously,

$$\bigl|f(t,u)\bigr|\leq\frac{e^{t}}{(t + 1)^{2}}.$$

Set

$$L_{2}=L_{3}=1,\qquad M_{2}=3,\qquad M_{3}=5 \quad\mbox{and}\quad \mu(t)=\frac{e^{t}}{(t + 1)^{2}}\in L^{4}\bigl([0, 1], R\bigr).$$

Thus, all the assumptions in Theorem 3.2 are satisfied. Hence, the impulsive fractional boundary value problem (4.2) has at least one solution on $$[0,1]$$.

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Acknowledgements

The authors express their sincere thanks to the anonymous reviews for their valuable suggestions and corrections for improving the quality of the paper. This work is supported by NSFC (11571207, 61503064), the Taishan Scholar project.

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Correspondence to Zhanbing Bai.

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